Find the area enclosed by the curve y=sinx and the x-axis between x=0 and x=22/7

To find the area enclosed by the curve y = sin(x) and the x-axis between x = 0 and x = 22/7, we can use integral calculus.

First, let's plot the curve y = sin(x) and the x-axis on a graph to visualize the region we are looking for.

The sine function (y = sin(x)) oscillates between -1 and 1. When it intersects the x-axis, y = 0.

Now, we want to find the area enclosed between the curve and the x-axis from x = 0 to x = 22/7 (or π).

To calculate this area, we need to find the definite integral of the absolute value of sin(x) with respect to x from 0 to 22/7.

∫[0, 22/7] |sin(x)| dx

To integrate the absolute value of sin(x), we need to break it up into two separate integrals based on where sin(x) is positive or negative.

∫[0, π] sin(x) dx + ∫[π, 22/7] -sin(x) dx

Now, we can find the antiderivative (integral) of sin(x) and -sin(x).

∫[0, π] sin(x) dx = -cos(x) evaluated from 0 to π

Substituting the values:

[-cos(π) - (-cos(0))] = [1 - (-1)] = 2

∫[π, 22/7] -sin(x) dx = cos(x) evaluated from π to 22/7

Substituting the values:

[cos(22/7) - cos(π)] = [cos(22/7) - (-1)] = [cos(22/7) + 1]

So, the enclosed area is 2 + [cos(22/7) + 1].

Using a calculator or mathematical software, you can approximate the cosine of 22/7 to get its numerical value. Add 2 to this value and you'll have the area enclosed by the curve y = sin(x) and the x-axis between x = 0 and x = 22/7.